Lemma: Let $E$ be a closed bounded subset of $\Omega$ and let $T$ be an injective transformation of class $C'$ on $\mathbb{R}^3$. Also the Jacobian does not vanish. Define $v(K)$ to be the volume of $K.$ Then $$\lim_{C\downarrow p} \frac{v(T(C))}{v(C)} = J(p)$$ where $C$ ranges over the family of cubes lying in $\Omega$ and having center $p$, and the limit is uniform for all $p\in E.$

In the proof of the above lemma, it says

If $p$ lies in the sphere $S$ with center $p_0$ and radius $r<\epsilon$, then $T(p)$ lies in the sphere whose center is $T(p_0)$ and whose radius is $(1+\epsilon)r.$ Moreover, when $p$ lies on the boundary of $S$, $T(p)$ lies outside the sphere with center $T(p_0)$ and with radius $(1-\epsilon)r.$

That stuff is fine and I understand the derivations which are in the book.

However, it then says

Since $T$ is injective and takes open sets into open sets, we see that $T(S)$ must contain the smaller sphere of radius $(1-\epsilon)r$.

I do not understand why injectivity and taking open sets to open sets implies that $T(S)$ contains the smaller sphere. That is where I am stuck.

Okay I guess I meant ball here. $T$ is a $C^1$ function, not necessarily linear. And I only want to know why the last line holds. This is part of a proof taken from Buck's Advanced Calculus.
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Clark KentJun 27 '11 at 13:55

@Glassjawed: I cannot speak for the downvoter, but it is tempting to downvote for lack of clarity. You can still edit your question to fix this. Buck's Advanced Calculus has many proofs, so saying that it is from that book doesn't help much. Will you please give a precise reference, and/or say precisely what it is a proof of?
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Jonas MeyerJun 28 '11 at 0:19

2 Answers
2

Not familiar with the book, but I'll answer just to stop this question from being rolled over by the hardworking Community hamster.

Compose $T$ with a linear transformation (the inverse of $T'(p_0)$) so that the composition, denoted $F$, satisfies $F'(p_0)=I$. Assuming we understand what Jacobians of linear maps have to do with volume, it remains to prove that $v(F(C))/v(C)\to 1$ as $C$ shrinks to $p_0$.

Since $F$ is $C^1$ smooth, for every $\epsilon>0$ there exists a neighborhood $N$ of $p_0$ in which $\|F'\|\le 1+\epsilon$. In this neighborhood $F$ is $(1+\epsilon)$-Lipschitz: that is, $|F(p)-F(q)|\le (1+\epsilon)|p-q|$ for any pair of points $p,q$.

The Lipschitz condition implies $v(F(C))\le (1+\epsilon)^n v(C)$ for any (measurable) set $C\subset N$. This follows from the definition of measure in terms of coverings. Or, for nice geometric shapes such as balls, this follows by a geometric argument based on containment.

Everything said in 2 and 3 applies to the inverse $F^{-1}$ as well, by the inverse function theorem. Hence $v(F(C))\ge (1+\epsilon)^{-n} v(C)$.

Warning: I'm using norm notation for distances (because it's easier to write) but the existence of a norm is not a prerequisite for this to work, just replace it with your distance metric if you'd like (which has to exist for "sphere" to be meaningful):